When you jump, you start by crouching down a bit, which lowers your center of mass to a distance d ≈ 40 cm below where it normally is. You then exert a downward contact force on the ground equal to about 2.5 times your weight (mg); we may assume that this force is constant while it acts. After your center of mass rises past its usual location, you leave the ground with an initial velocity v, and you reach a maximum height h. Ignore air resistance for this problem.
a) Draw a free-body diagram for your body during the interval while you are pushing against the ground with a force of 2.5mg. What is the net force on your body during that time?
b) What is the work done on the Earth by the force of your feet pushing on the Earth during that time?
c) What is the total work done by gravity on your body during the time while you are pushing against the ground? What is the total work done by the normal force on your body during the time while you are pushing against the ground?
d) Find the initial kinetic energy of your body as it leaves the ground using the Work Kinetic Energy theorem and your answers to part c). What is your takeoff speed v? How high will you jump (h)?
Here the downward force when you are crouching is mg (weight). Normal acts on your body in upward direction.
During the time when you are pushing the ground with 2.5mg,
N - mg = 2.5mg or N = 3.5mg ; Net force = 2.5 mg
Now when you apply the force 2.5mg on the ground, an equal and opposite force acts on your body in upward direction responsible for the jump.
b) While you are pushing the ground, the force acts but the point of application does not move. So, no work is done.
c) Work done by gravity = mg * 0.4 = 0.4mg
Work done by normal = 3.5mg * -0.4 = -1.4 mg
The negative work shows that the energy is stored in the form of potential which when released will be responsible for the jump.
d) when the body leaves the ground the CoM is again at the normal height and the work done is 1.4mg - 0.4 mg = mg
This was the energy stored as the potential energy in the body.
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